3.649 \(\int \cos ^2(c+d x) (a+b \tan (c+d x))^n \, dx\)

Optimal. Leaf size=272 \[ -\frac {\left (\sqrt {-b^2} \left (\frac {a^2}{b^2}-n+1\right )-a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{4 b d (n+1) \left (\frac {a^2}{b^2}+1\right ) \left (a-\sqrt {-b^2}\right )}+\frac {b \left (\sqrt {-b^2} \left (\frac {a^2}{b^2}-n+1\right )+a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{4 d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right )}+\frac {\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

[Out]

-1/4*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)))*(-a*n+(1+a^2/b^2-n)*(-b^2)^(1/2))*(a+b*tan(d*
x+c))^(1+n)/(1+a^2/b^2)/b/d/(1+n)/(a-(-b^2)^(1/2))+1/4*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(
1/2)))*(a*n+(1+a^2/b^2-n)*(-b^2)^(1/2))*(a+b*tan(d*x+c))^(1+n)/(a^2+b^2)/d/(1+n)/(a+(-b^2)^(1/2))+1/2*cos(d*x+
c)^2*(b+a*tan(d*x+c))*(a+b*tan(d*x+c))^(1+n)/(a^2+b^2)/d

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Rubi [A]  time = 0.46, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3506, 741, 831, 68} \[ -\frac {\left (\sqrt {-b^2} \left (\frac {a^2}{b^2}-n+1\right )-a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{4 b d (n+1) \left (\frac {a^2}{b^2}+1\right ) \left (a-\sqrt {-b^2}\right )}+\frac {b \left (\sqrt {-b^2} \left (\frac {a^2}{b^2}-n+1\right )+a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{4 d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right )}+\frac {\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Tan[c + d*x])^n,x]

[Out]

-((Sqrt[-b^2]*(1 + a^2/b^2 - n) - a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]
)]*(a + b*Tan[c + d*x])^(1 + n))/(4*(1 + a^2/b^2)*b*(a - Sqrt[-b^2])*d*(1 + n)) + (b*(Sqrt[-b^2]*(1 + a^2/b^2
- n) + a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1
+ n))/(4*(a^2 + b^2)*(a + Sqrt[-b^2])*d*(1 + n)) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x])*(a + b*Tan[c + d*x])^(
1 + n))/(2*(a^2 + b^2)*d)

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{\left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^n \left (-1-\frac {a^2}{b^2}+n+\frac {a n x}{b^2}\right )}{1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \left (\frac {\left (-a n+\sqrt {-b^2} \left (-1-\frac {a^2}{b^2}+n\right )\right ) (a+x)^n}{2 \left (\sqrt {-b^2}-x\right )}+\frac {\left (a n+\sqrt {-b^2} \left (-1-\frac {a^2}{b^2}+n\right )\right ) (a+x)^n}{2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}+\frac {\left (b \left (\sqrt {-b^2} \left (1+\frac {a^2}{b^2}-n\right )-a n\right )\right ) \operatorname {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {\left (b \left (\sqrt {-b^2} \left (1+\frac {a^2}{b^2}-n\right )+a n\right )\right ) \operatorname {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}\\ &=-\frac {b \left (\sqrt {-b^2} \left (1+\frac {a^2}{b^2}-n\right )-a n\right ) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) \left (a-\sqrt {-b^2}\right ) d (1+n)}+\frac {b \left (\sqrt {-b^2} \left (1+\frac {a^2}{b^2}-n\right )+a n\right ) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right ) d (1+n)}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 1.27, size = 225, normalized size = 0.83 \[ \frac {(a+b \tan (c+d x))^{n+1} \left (-\frac {\left (\sqrt {-b^2} \left (a^2-b^2 (n-1)\right )-a b^2 n\right ) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{(n+1) \left (a-\sqrt {-b^2}\right )}+\frac {\left (a^2 \sqrt {-b^2}+a b^2 n+\left (-b^2\right )^{3/2} (n-1)\right ) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{(n+1) \left (a+\sqrt {-b^2}\right )}+2 b \cos ^2(c+d x) (a \tan (c+d x)+b)\right )}{4 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Tan[c + d*x])^n,x]

[Out]

((a + b*Tan[c + d*x])^(1 + n)*(-(((Sqrt[-b^2]*(a^2 - b^2*(-1 + n)) - a*b^2*n)*Hypergeometric2F1[1, 1 + n, 2 +
n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])])/((a - Sqrt[-b^2])*(1 + n))) + ((a^2*Sqrt[-b^2] + (-b^2)^(3/2)*(-1 +
n) + a*b^2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])])/((a + Sqrt[-b^2])*(1
+ n)) + 2*b*Cos[c + d*x]^2*(b + a*Tan[c + d*x])))/(4*b*(a^2 + b^2)*d)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*cos(d*x + c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*cos(d*x + c)^2, x)

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maple [F]  time = 1.18, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

[Out]

int(cos(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*cos(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + b*tan(c + d*x))^n,x)

[Out]

int(cos(c + d*x)^2*(a + b*tan(c + d*x))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \cos ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*cos(c + d*x)**2, x)

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